Question

If a,b,c are in A.P. ; b,c,d are in G.P & \(\frac{1}{c}\),\(\frac{1}{d}\),\(\frac{1}{e}\) are in A.P prove that a,c,e are in G.P.

Answer 1

@ganeshie8

Answer 2

@priyar

Answer 3

Maybe turn the G.Ps into equations? \[
\frac cb = \frac dc
\]

Answer 4

\[
b-a = c-b
\]And \[
\frac 1d - \frac 1c = \frac 1e -\frac 1d
\]

Answer 5

@wio plez tell from which eqn i should start ?

Answer 6

i think \[\frac{c-d}{b-c}=\frac{c+d}{b+c}\]

Answer 7

Let the AP be \(a, a+k, a+2k\). Now given the second statement\[d = \frac{(a+2k)^2}{a+k}\]Also\[\frac{1}e = \frac{2}d - \frac{1}c = \frac{2(a+k)}{(a+2k)^2} - \frac{1}{a+2k}=\frac{a}{(a+2k)^2}\]So\[a = a\]\[c = a+2k\]\[e = \frac{(a+2k)^2}a\]Do you see it now? :)

Answer 8

@ParthKohli how did u gt \[d=\frac{(a+2k)^2}{(a+k)}\]

Answer 9

\(a_1, a_2, a_3\) in GP:\[a_2^2 = a_1 a_3\]or\[a_3 = \frac{a_2^2}{a_1}\]

Answer 10

Thanx @ParthKohli now i gt it :)